I was reading Feynman lectures III's Spin One; there at the machinery of quantum mechanics he discusses a situation in which he needs to find the amplitude of finding the particle at ${\chi}$ state after it travels through an apparatus $A$ before being at state $\phi$. So, the amplitude can be written as $\langle \chi|A|\phi\rangle$.

But, there it was not done so; Feynman reasoned that:

[...]we would want to know the amplitude $$\langle \chi|A|\phi\rangle$$ A complete analysis of the apparatus $A$ would have to give the amplitude $\langle \chi|A|\phi\rangle$ for every possible pair of states $\phi$ and $\chi$— an infinite number of combinations!

Now, $\phi$ only contains three base states $+S$,$0$,$-S$; $\chi$ contains three base states $+R$,$0$,$-R$; in the middle exits the apparatus $A$ - how can the three states on the left & three states on the right along with $A$ make "infinite number of combinations!"?

How then can we give a concise description of the behavior of the apparatus A? We can do it in the following way. Imagine that the apparatus of (5.28) is modified to be

enter image description here

This is really no modification at all since the wide-open T apparatuses don’t do anything. But they do suggest how we can analyze the problem. There is a certain set of amplitudes $\langle i|+S\rangle$ that the atoms from $S$ will get into the $i$ state of $T$. Then there is another set of amplitudes that an $i$ state (with respect to $T$) entering $A$ will come out as a $j$ state (with respect to $T$). And finally there is an amplitude that each $j$ state will get through the last filter as a ($0R$) state. For each possible alternative path, there is an amplitude of the form $$\langle 0R|j\rangle\langle j|A| i\rangle\langle i| +S\rangle$$and the total amplitude is the sum of the terms we can get with all possible combinations of $i$ and $j$. The amplitude we want is $$\sum_{ij} \langle 0R|j\rangle\langle j|A| i\rangle\langle i| +S\rangle$$.

If ($0R$) and ($+S$) are replaced by general states $χ$ and $ϕ$, we would have the same kind of expression; so we have the general result $$\langle \chi|A|\phi\rangle = \sum_{ij} \langle \chi|j\rangle\langle j|A| i\rangle\langle i| \phi\rangle$$

Now notice that the right-hand side of Eq. (5.32) is really “simpler” than the left-hand side. The apparatus A is completely described by the nine numbers $\langle j|A|i\rangle$ which tell the response of $A$ with respect to the three base states of the apparatus $T$. Once we know these nine numbers, we can handle any two incoming and outgoing states $ϕ$ and $χ$ if we define each in terms of the three amplitudes for going into, or from, each of the three base states.

This then is the machinery of quantum mechanics for a spin-one particle. Every state is described by three numbers which are the amplitudes to be in each of some selected set of base states. Every apparatus is described by nine numbers which are the amplitudes to go from one base state to another in the apparatus. From these numbers anything can be calculated.

Why is $\langle \chi|j\rangle\langle j|A| i\rangle\langle i| \phi\rangle$ "simpler" than $\langle \chi|A|\phi\rangle$ ? How did by introduction of $T$, it became "simpler"?

I am not getting what problem there would if we simply measure $\langle \chi|A|\phi\rangle$; how does it arise infinite number of combinations, after all the sets of $\phi$ & $\chi$ contain only three states each? Can anyone please help me explain how three base states give rise to infinite number of combinations? And also, what advantage do we get if we introduce the open-channel apparatus $T$??

  • $\begingroup$ I think I should notify you that you haven't defined \SGP etc in $\LaTeX$ hence MathJax doesn't render. $\endgroup$
    – Gonenc
    Jul 29, 2015 at 9:36
  • $\begingroup$ Couldn't you draw it on paper and take a photo? Or draw it on paint. $\endgroup$
    – jinawee
    Jul 29, 2015 at 9:52
  • $\begingroup$ full dissclosure: i didnt read your text to full detail. The main point seems to be why it might be advantageous to twice "insert unity". This way you have to compute the "matrix elements" of A in a basis that you chose. If you clever you do this using states i,j where you know what happens if you apply A $\endgroup$
    – Bort
    Jul 29, 2015 at 10:09
  • 1
    $\begingroup$ @user36790: I don't know what \SPG stand for but you can always define a new command with \newcommand{name goes here}[number of parameters]{and description} Feel free to check the wiki page for more information. $\endgroup$
    – Gonenc
    Jul 29, 2015 at 10:12

2 Answers 2


So let's say we have the three base states $\lvert + \rangle$, $\lvert 0\rangle$, $\lvert - \rangle$. A general state is then: $$\lvert \psi \rangle = c_+ \lvert + \rangle + c_0 \lvert 0\rangle + c_- \lvert - \rangle$$ where, for normalization, $\lvert c_+\rvert^2 + \lvert c_0\rvert^2 + \lvert c_-\rvert^2 = 1$. You can clearly make an infinite number of choices for the $c_i$s, and so there are an infinite number of states that you can make from these base states.

It is important to note that in quantum mechanics, it is of physical significance not only which base states are combined, but exactly how they are combined (i.e. the coefficients). Apart from the information of probabilities, the coefficients also give what the system looks like in another base, and so the probabilities for different bases. And each combination of coefficients produces a completely different physical state.

So yes, instead of specifying infinite $\lvert \psi \rangle$s, we can simply specify the action of $A$ on the base states. This requires us to know only 3 vectors (or equivalently, nine numbers - the expression of each of the 3 vectors in the basis): $A\lvert + \rangle$, $A\lvert 0 \rangle$, $A\lvert - \rangle$ (correspondingly, all the $\langle j \rvert A\lvert i \rangle$).Thanks to the linearity of quantum mechanics, the action of $A$ on $\lvert \psi \rangle$ is then simply: $$A\lvert \psi \rangle = c_+ A\lvert + \rangle + c_0 A\lvert 0\rangle + c_- A\lvert - \rangle$$ There you have it - a complete description of $A$ would do to a general state (which is one among an infinite number of possible states), in terms of nine numbers, all made possible by expanding in the basis.

To recap: expanding the general state in a basis helps us to explain the action of the "appartus" on the general state, simply in terms of the action on each vector in the basis, which is much easier to keep track of.

This is a lot like writing the dot product $\mathbf{a}\cdot\mathbf{b}$ for 3D vectors $$\mathbf{a} = a_x\mathbf{e_x} + a_y\mathbf{e_y} + a_z\mathbf{e_z}$$ $$\mathbf{b} = b_x\mathbf{e_x} + b_y\mathbf{e_y} + b_z\mathbf{e_z}$$ as: $$\mathbf{a}\cdot\mathbf{b} = a_x b_x + a_y b_y + a_z b_z$$ just using linearity and the extremely simple known products: $$\mathbf{e_i} \cdot \mathbf{e_j} = \delta_{ij}$$ Without these, we'll have to specify $\mathbf{a}\cdot\mathbf{b}$ for any pair of vectors, and there is an infinite number of such pairs.

  • $\begingroup$ Hi, sir! It was a nice answer but I fear that it is not actually directly answering the questions in the sense I asked(sorry:|). (Have you mingled the $+$ state of $T$ with that of $S$? Caution: They are different:) ... Nevertheless, it has helped me by a great extent to understand the lectures a bit more:) +1. $\endgroup$
    – user36790
    Jul 30, 2015 at 3:08
  • $\begingroup$ @user36790 In the expansion for $\langle \chi \rvert A\lvert \phi \rangle$, it is only the $T$ basis that is considered by Feynman. The equation prior to that is an example in terms of the pure $R$ and $S$ base states, which is not quite the general state. The idea is you need to expand these base states in the $T$ basis as well to use the elements of $A$ in the $T$ basis, and this expansion is conceptually the same as expanding the general state. $\endgroup$
    – AV23
    Jul 30, 2015 at 9:24
  • $\begingroup$ I am now briefing what I have conceived: in order to analyze the apparatus $A$, we have to find all the amplitudes $\langle \chi |A| \phi\rangle$ for each $\chi$ & $\phi$ which can be infinite (at least many to get bothered). So, in order to analyze the apparatus $A$, $T$ open channel apparatuses have been introduced so that only nine numbers are required to analyze $A$ immaterial of $\phi$ & $\chi$ . Right? $\endgroup$
    – user36790
    Jul 30, 2015 at 9:31
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    $\begingroup$ @user36790 mostly right. But "at least many to get bothered" is not quite accurate - it actually is infinite. Think of the infinite number of directions in a 3D vector space. $\endgroup$
    – AV23
    Jul 30, 2015 at 14:45

So, the context is, "We want to show you why these laws are useful." He has just mentioned three laws, and he wants to prove that they constrain any apparatus to be a linear (really, unitary) operator on the Hilbert space.

Feynman's notation is a little sloppy because these were meant to be presented in lecture, so let me give a hopefully more clear presentation of the argument with a more-agnostic notation.

Functions get linearized

Feynman is here saying: suppose you want to describe the apparatus $\mathcal A$, which in quantum mechanics is some unitary operator. If we just look at it as a mapping $\mathcal A: \mathbb C^3 \to \mathbb C^3$, then we in theory have three functions of three continuous arguments to deal with, $\phi'_{1,2,3}(\phi_1,\phi_2,\phi_3).$ If you don't seem to have any other information, and you're describing some arbitrary function, you need an infinite amount of information for any one of these: That's no good. Let's write the effect of the arbitrary apparatus on the wavefunction variously as $\mathcal A\big[|\phi\rangle\big] = |\mathcal A(\phi)\rangle$.

Now the laws of quantum mechanics come in and say, "hey, you can always insert the identity operator $I = \sum_k |k\rangle\langle k|$ into the middle of any expression." They also say, from how they construct the amplitudes, "hey, $\mathcal A$ is linear: both $\mathcal A \big[|a\rangle + |b\rangle\big] = |\mathcal A(a)\rangle + |\mathcal A(b)\rangle$ and $\mathcal A\big[c |\psi\rangle\big] = c ~ |\mathcal A(\psi)\rangle$ for any complex number $c$."

As a consequence of these two directly we can write:$$\langle\chi|\mathcal A(\phi)\rangle = \sum_{\{i,j\}}\langle\chi|j\rangle\langle j|~~\mathcal A\Big[|i\rangle ~\langle i | \phi\rangle \Big]$$If we write $\langle i | \phi\rangle = \phi_i$ and similarly for $|\chi\rangle$, those are pure complex numbers and we can see that the result is given by$$\langle\chi|\mathcal A(\phi)\rangle = \sum_{\{i,j\}}~\chi^*_j ~ \phi_i ~~\langle j | \mathcal A(i)\rangle$$ We can regard this last term as simply a set of 9 numbers $A_{ji}$, whence we discover that in fact:$$\phi_k'(\phi_1, \phi_2, \phi_3) = A_{k1} ~\phi_1 + A_{k2} ~\phi_2 + A_{k3} ~\phi_3$$If $\mathcal A$ didn't have this form then either we couldn't insert the identity matrix wherever we pleased, or we wouldn't be able to calculate amplitudes the way that we do, just assuming that the functions are linear.

The hard work is being done really by linearity but the finite basis certainly helps even for describing $|\phi\rangle$.

Extra: $A$ is unitary to preserve probability

In fact, if $A^\dagger$ is the conjugate transpose of this matrix, then we also need, to preserve probability, that $A^\dagger~A = A~A^\dagger = 1.$ The diagonal is easy to see: $$1 = \langle \mathcal A(i) |\mathcal A(i)\rangle = \sum_j \Big| \langle j|\mathcal A(i)\rangle \Big|^2 = \sum_j (A_{ji})^* A_{ji} .$$That the off-elements have to be 0 can be seen in many ways; the easiest is to define $|i, k, \theta \rangle = \sqrt{1\over 2} |i\rangle + \sqrt{1\over 2} e^{i\theta} |k\rangle $ and then examine $$1 = \langle \mathcal A(i,k,\theta) |\mathcal A(i,k,\theta)\rangle = 1 + {e^{i\theta}\over 2} \langle \mathcal A(i)|\mathcal A(k)\rangle + {e^{-i\theta}\over 2}\langle \mathcal A(k)|\mathcal A(i)\rangle$$For $\theta = 0$ we find that the real part is 0; for $\theta = \pi/2$ we find that the imaginary part is too. The only escape from the latter argument happens is if the input vector doesn't have length 1, which can only happen if $i = k$ due to the orthonormality of the basis.


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